Question

A horizontal wire of length 3.0 m carries a current of 6.0 A and is oriented so that the current direction is 50° S of W. The Earth's magnetic field is due north at this point and has a strength of 0.14 ´ 10-4 T. What is the size of the force on the wire?

Answer 1

Explanation:

tjng

hr6.hr

jtsjtjtzhtjzjtsjtstztsjm

mgzjgxjtx

Answer 2

Here, I=5A;1=0.1m;

Mass per unit length of the wire

λ=3×10

−3

kgm

−1

Therefore, weight of the wire,

mg=3×10

−3

×0.1×9.8=2.94×10

−3

N

Let B be the strength of the magnetic field applied.

F=BIl=B×5×0.1=0.5B

In equilibrium,

F=mg

∴0.5B=2.94×10

−3

B=

0.5

2.94×10

−3

=5.88×10

−3

T